Since $1+x^2>0$ for all real $x$, divide the equation by that factor. Its linear coefficient can be integrated by polynomial division:Thus a positive integrating factor is $M=e^{3x^2/2}(1+x^2{)}^{-3/2}$. Multiplying the divided equation gives the product rule formThe integration by substitution $u=1+x^2$ givesThe initial condition $y(0)=1$ fixes $C-2=1$. The required solution is thereforeNo real singularity was introduced by the division or the integrating factor, so the solution extends over the entire real line.

For the first-order linear ordinary differential equation, seek an integrating factor $M(x)$ that makes its left side a product rule derivative. Multiplying by $M$ gives $M{y}^{\prime }+MPy$, and comparison with $(My{)}^{\prime }=M{y}^{\prime }+M^{\prime }y$ requires $M^{\prime }=PM$. On an interval where $P$ is continuous chooseThis factor is never zero. The transformed equation is $(My{)}^{\prime }=MQ$, so integration from $x_0$ to $x$ yieldsHere $C=y(x_0)$ for an initial value problem. The indefinite integral version is $y=e^{-\int Pdx}[C+\int Q{e}^{\int Pdx}dx]$, with any constant choices inside the integrating factor absorbed into $C$. Continuity of $P,Q$ is a convenient sufficient hypothesis for a classical differentiable solution of a differential equation.

The homogeneous characteristic equation is $r^2+2r=r(r+2)=0$, giving the complementary solution $C_1+C_2{e}^{-2x}$. For a particular solution, try $A{x}^2+Bx$. Substitution gives $2A+4Ax+2B=4x$, so $A=1$ and $B=-1$. Therefore the general solution isAs a direct check, the polynomial part contributes $2+2(2x-1)=4x$, while both complementary terms contribute zero.

Set $w=e^z\ne 0$. The hyperbolic cosine equation becomes $w+w^{-1}=-4$, hence $w^2+4w+1=0$. Its two polynomial roots are $w=-2\pm \sqrt{3}$, both negative. Taking every value of the complex logarithm yieldsSince $(2-\sqrt{3})(2+\sqrt{3})=1$, their real logarithms have opposite signs. All solutions are thereforeThe positive real constant is also $\mathrm{arcosh}2$, the inverse hyperbolic cosine of $2$. Substitution gives $\cosh z=-\cosh (\ln (2+\sqrt{3}))=-2$ for both signs and every integer $k$.

A constant real part $\mathrm{Re}\ln z=L$ means $\ln |z|=L$, so its level sets are the concentric circlesTheir radii are strictly positive. The real part is independent of the logarithm's branch, so there is no branch ambiguity in these circles.

For $\mathrm{Re}(1/z)=K$, the equation is $x=K(x^2+y^2)$. If $K\ne 0$, complete the square to obtainThese circles have centres on the real axis and are tangent to the imaginary axis at the excluded origin. Positive $K$ puts the circle to the right, negative $K$ to the left. The separate case $K=0$ gives the punctured imaginary axis $\boxed{x=0,y\ne 0}$. The sketch uses an open marker at the origin to emphasize that $1/z$ is never defined there.

In polar coordinates, $x=r\cos \theta$, $y=r\sin \theta$, with $r=\sqrt{x^2+y^2}$. Thus the polar form of a complex number isHere $r=|z|\ge 0$ is its complex modulus; for $z\ne 0$, the complex argument $\theta$ is determined modulo $2\pi$. At $z=0$ the angle is undefined, although every value of $\theta$ gives the same point when $r=0$.

The dot product formula for the angle between two nonzero vectors gives

The Euclidean norm of each position vector gives

The separable differential equation gives $dy/y=-dx$, so $y=C{e}^{-x}$. The initial condition fixes $C=2$, and hence

On the negative half-line, an antiderivative of $1/x$ is $\ln |x|$. The fundamental theorem of calculus therefore givesThe negative sign agrees with the negative integrand throughout the interval.